# Eigenvalues of differential operator

If $L : C^2[a,b] \rightarrow C^0[a,b]$ is s.t. $L y(t) = \ddot y(t) +p \dot y + q y(t)$ and $L$ is invertible then $L^{-1}$ has at most countable eigenvalues and they accumulate in $0$.

Why countable? And why should they accumulate in $0$?

• As stated, I don't think L can be invertible, since for any $y(a), y'(a)$ there is a unique $y(t)$ with $Ly = 0$. – Robert Lewis Sep 27 '13 at 1:37